What is the value of $\dfrac{d}{dx}\left(\sqrt[5]{x^2}\right)$ at $x=32$ ?
The strategy We can first rewrite the radical as a rational power of $x$. Then, the derivative can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Once we have the derivative, we can evaluate it at $x=32$. Rewriting the radical as a rational power $\sqrt[5]{x^2}=x^{^{\frac{2}{5}}}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(x^{^{\frac{2}{5}}}\right) \\\\ &=\dfrac{2}{5}x^{^{\frac{2}{5}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac25x^{^{-\frac{3}{5}}} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\sqrt[5]{x^2}\right)=\dfrac25x^{^{-\frac{3}{5}}}$, which can also be written as $\dfrac{2}{5(\sqrt[5]{x})^3}$. Now let's plug ${x=32}$ : $\begin{aligned} \dfrac{2}{5\left(\sqrt[5]{{32}}\right)^3}&=\dfrac{2}{5(2)^3} \\\\ &=\dfrac{2}{5\cdot 8} \\\\ &=\dfrac{1}{20} \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\sqrt[5]{x^2}\right)$ at $x=32$ is $\dfrac{1}{20}$.